This problem is on Di Francesco's book I. It's exercise 7.1:
Calculate the norm of the following vector, where $\lvert h\rangle$ is the state of highest weight.

$$L_{-1}^n\lvert h\rangle$$

I have just tried to use the commutation relations of the $L$ operators and the fact that $L_1$ acts on $\lvert h\rangle$ is $0$. But as the calculation goes on, things began to be troublesome. I just found them too complicated.

Commutation relations:
$$[L_n,L_m]=(n-m)L_{n+m}+\frac{c}{12}\delta_{n+m,0}n(n^2-1) $$
It is in fact asking us to calculate:
$\langle h|L_1^nL_{-1}^n|h\rangle $.
And we have the following relations:
$$\langle h\rvert L_{-1}=0 \quad\mbox{and}\quad L_1\lvert h\rangle=0. $$

This is a legitimate and very high-level technical questions, and I would bet quite a large amount of money that the closevoter(s) here have no clue what the OP is talking about. And just to say, even Qmechanic deems it worth an answer, so voting to close on such questions is quite preposterous ...
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DilatonFeb 23 '14 at 18:19

I'll try again later this week, given you said that.
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Li XingheFeb 24 '14 at 14:54

This problem really intrigues me. I won't see the answer or hint before another try.
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Li XingheFeb 24 '14 at 14:55

ii) Using the fact that $L_1 L_{-1}=[L_1, L_{-1}]+L_{-1} L_{1}$ and that $L_0|h\rangle=h|h\rangle $, try to calculate $\langle h|L_1 L_{-1}|h\rangle =\cdots\stackrel{?}{=}2h$.

iii) Calculate the following quantity that will (probably) be useful later:
$$\langle h|(L_1 L_{-1})^n|h\rangle =\cdots\stackrel{?}{=}(2h)^n$$

iv) (The hard part) We have done the case for $n=1$ for the quantity $\langle h |L_1^n L_{-1}^n |h\rangle$ in i), but you will also need to do the case for $n=2$ and $n=3$ by hand, using the formulae in i) and ii) above and from that, try to deduce an inductive formula for the general case. I will have to admit though that deducing the general form (which then can be proved by induction) from a few cases might be quite hard!